How do you solve #sin(x+60°) + cos(x+30°) = 1/2# in the range 0° < x < 360°?

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Answer 1 · 2015-05-02T03:35:55

# x = 73.22^@# and #x = 360 - 73.22 = 286.78^@#

Use the trig identities:
#sin (a + b) = sin a.cos b + sin b.cos a#
#cos (a + b) = cos a.cos b - sin a.sin b#

#sin (x + 60) = (1/2).sin x + ((sqrt3)/2).cos x #

#cos (x + 30) = cos x.((sqrt3)/2) - (1/2).sin x#

Adding the 2 expressions, we get:

#(sqrt3).cos x= 1/2 -> cos x = 1/(2sqrt3) = (sqrt3)/6#

-># x = 73.22^@ # and #x = 360 - 73.22 = 286.78^@#